Q.Let A, B, C, D be any four points in space. Prove that $|\overline{AB}\times\overline{CD}+\overline{BC}\times\overline{AD}+\overline{CA}\times\overline{BD}|=4(\text{area of }\triangle ABC)$.

Let $A,B,C,D$ have position vectors $\bar a,\bar b,\bar c,\bar d$. Compute each term:

Expanding all three using bilinearity, every term that involves $\bar d$ actually cancels out completely

across the three expansions (this is because $D$'s position vector $\bar d$ appears once in each term but the

coefficients of the "other" vector it's paired with, summed across all three products, telescope to zero --

a direct but lengthy expansion confirms this), leaving only terms in $\bar a,\bar b,\bar c$:

(the factor of 2 emerges from collecting the surviving like terms in the full expansion). Taking magnitudes and …

Expand all three cross products in terms of position vectors using bilinearity, observe that every term containing D's position vector cancels across the three expansions (a remarkable fact that makes the whole expression independe …

Assuming the identity must depend on D's location and trying to introduce extra assumptions about where D is, rather than pushing through the alg …